The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic,when ($K$ is a positive constant)

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

In the following table, the displacement of a $Simple \text{ } Harmonic \text{ } Oscillator \text{ } (SHO)$ is shown in column-$I$ and the kinetic energy is shown in column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

$A$ particle is executing simple harmonic motion with a time period of $3 \,s$. At a position where the displacement of the particle is $60 \%$ of its amplitude, the ratio of the kinetic and potential energies of the particle is

$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.

Show that for a particle in linear $SHM$,the average kinetic energy over a period of oscillation equals the average potential energy over the same period.